Biostat 212a Homework 4

Due Mar. 5, 2024 @ 11:59PM

1 ISL Exercise 8.4.3 (10pts)

3. Consider the Gini index, classification error, and entropy in a simple classification setting with two classes. Create a single plot that dis- plays each of these quantities as a function of pˆm1. The x-axis should display pˆm1, ranging from 0 to 1, and the y-axis should display the value of the Gini index, classification error, and entropy. Hint: In a setting with two classes, pˆm1 = 1 − pˆm2. You could make this plot by hand, but it will be much easier to make in R.

p1 <- seq(0, 1, 0.01)
p2 <- 1 - p1
gini <- 2 * p1 * p2
class.error <- 1 - pmax(p1, p2)
entropy <- -pmax(p1, p2) * log2(pmax(p1, p2)) - pmin(p1, p2) * log2(pmin(p1, p2))
matplot(p1, cbind(gini, class.error, entropy), ylab = "Gini index, Classification error, Entropy", type = "l", col = c("green", "blue", "orange"))
legend("topright",legend=c("Gini index","Class.error", "Entropy"),pch=20,col=c("green", "blue", "orange"))

2 ISL Exercise 8.4.4 (10pts)

X2 < 1 | 15 5 3 0 10 X2 1 0 01 X1 X1 < 0 X1 < 1 X2 < 2 0.21 2.49 -1.80 0.63 -1.06 FIGURE 8.14. Left: A partition of the predictor space corresponding to Exer- cise 4a. Right: A tree corresponding to Exercise 4b. This question relates to the plots in Figure 8.14. (a) Sketch the tree corresponding to the partition of the predictor space illustrated in the left-hand panel of Figure 8.14. The num- bers inside the boxes indicate the mean of Y within each region. If X1≥1 then 5, else if X2≥1 then 15, else if X1<0 then 3, else if X2<0 then 10, else 0.

library(knitr)
include_graphics("/Users/yangan/Desktop/212A/212a-hw/hw4/8.4.4(a).jpg")

2. Create a diagram similar to the left-hand panel of Figure 8.14, using the tree illustrated in the right-hand panel of the same figure. You should divide up the predictor space into the correct regions, and indicate the mean for each region.

# (b)
par(xpd = NA)
plot(NA, NA, type = "n", xlim = c(-2, 2), ylim = c(-3, 3), xlab = "X1", ylab = "X2")
# X2 < 1
lines(x = c(-2, 2), y = c(1, 1))
# X1 < 1 with X2 < 1
lines(x = c(1, 1), y = c(-3, 1))
text(x = (-2 + 1)/2, y = -1, labels = c(-1.8))
text(x = 1.5, y = -1, labels = c(0.63))
# X2 < 2 with X2 >= 1
lines(x = c(-2, 2), y = c(2, 2))
text(x = 0, y = 2.5, labels = c(2.49))
# X1 < 0 with X2<2 and X2>=1
lines(x = c(0, 0), y = c(1, 2))
text(x = -1, y = 1.5, labels = c(-1.06))
text(x = 1, y = 1.5, labels = c(0.21))

3 ISL Exercise 8.4.5 (10pts)

5. Suppose we produce ten bootstrapped samples from a data set containing red and green classes. We then apply a classification tree to each bootstrapped sample and, for a specific value of X, produce 10 estimates of P(Class is Red|X): 0.1,0.15,0.2,0.2,0.55,0.6,0.6,0.65,0.7, and0.75. There are two common ways to combine these results together into a single class prediction. One is the majority vote approach discussed in this chapter. The second approach is to classify based on the average probability. In this example, what is the final classification under each of these two approaches?

p <- c(0.1, 0.15, 0.2, 0.2, 0.55, 0.6, 0.6, 0.65, 0.7, 0.75)
# Average probability
mean(p)

[1] 0.45

In this case, the most common prediction is 0.6, which occurs twice. With the majority vote approach, we classify X as Red as it is the most commonly occurring class among the 10 predictions (6 for Red vs 4 for Green). With the average probability approach, we classify X as Green as the average of the 10 probabilities is 0.45.

4 ISL Lab 8.3. Boston data set (30pts)

Follow the machine learning workflow to train regression tree, random forest, and boosting methods for predicting medv. Evaluate out-of-sample performance on a test set.

5 ISL Lab 8.3 Carseats data set (30pts)

Follow the machine learning workflow to train classification tree, random forest, and boosting methods for classifying Sales <= 8 versus Sales > 8. Evaluate out-of-sample performance on a test set.